| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring substitution into simultaneous equations to find constants, differentiation for acceleration, solving a simple equation for time at rest, and integration for displacement. All techniques are standard M1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4b + 4^{\frac{3}{2}}c = 8\ [\to 4b + 8c = 8]\) and \(9b + 9^{\frac{3}{2}}c = 13.5\ [\to 9b + 27c = 13.5]\); State \(b = 3\) and \(c = -0.5\). OR \(3\times4 + (-0.5)\times4^{\frac{3}{2}} = 8\) AND \(3\times9 + (-0.5)\times9^{\frac{3}{2}} = 13.5\) | B1 | Must have 2 correct equations, which do not have to be simplified. Allow to just state the values of \(b\) and \(c\). Allow substitution of \(b=3\) and \(c=-0.5\) in both equations to verify. No further calculation required. B0 if any incorrect work seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(a = \dfrac{dv}{dt} =\right) 3 - 0.5\times\dfrac{3}{2}\times t^{\frac{1}{2}}\) OR \(b + c\times\dfrac{3}{2}\times t^{\frac{1}{2}}\) | M1 | Attempt to differentiate, decrease power by 1 and a change in coefficient in at least one term (which must be the same term); allow unsimplified; \(a = \dfrac{v}{t}\) is M0. |
| acceleration \(= 2.25\ \text{ms}^{-2}\) | A1 | OE, e.g. \(\dfrac{9}{4}\) or \(2\dfrac{1}{4}\). SC B1 for 2.25 if no differentiation seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate \(v\) to 0 and attempt to solve for \(t\) | M1 | Must get to \(t = \ldots\) and must be positive. |
| \(t = 36\) ONLY | A1 | WWW. Allow if \(t = 0\) seen and not rejected. |
| Attempt to integrate | M1 | Increase power by 1 and a change in coefficient in at least one term (which must be the same term). \(s = vt\) is M0. |
| \(s = \dfrac{3}{2}t^2 - 0.5\times\dfrac{2}{5}\times t^{\frac{5}{2}}(+D) = \dfrac{3}{2}t^2 - \dfrac{1}{5}t^{\frac{5}{2}}(+D)\) OR \(s = \dfrac{b}{2}t^2 + c\times\dfrac{2}{5}\times t^{\frac{5}{2}}(+D)\) | A1 | Allow unsimplified (including indices). |
| Sub \(t = 36\) (or use limits 36 and 0) to get distance \(= 388.8\) m ONLY | A1 | Allow 389 m. If no integration seen for the last 3 marks, allow SC B1 for 388.8 m. Max M1A1B1 for 3/5 marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{3}{2}t^2 - 0.5\times\dfrac{2}{5}\times t^{\frac{5}{2}} = 0\) | M1 | Equate *their* \(s\) (that has come from an integration attempt) to 0 and attempt to solve for \(t\). Must get to \(t = \ldots\) |
| \(t = 56.25\) | A1 | WWW. OE, e.g. \(\dfrac{225}{4}\). Allow 2sf or better. Allow any correct unsimplified expression equivalent to \(56.25\) e.g. \(\dfrac{15^2}{2^2}\) or \(7.5^2\). |
| Speed \(= 42.2\ \text{ms}^{-1}\) ONLY | A1 | AWRT 42.2. Speed \(= -42.2\) is A0. Allow A1 if negative sign dropped without justification. |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4b + 4^{\frac{3}{2}}c = 8\ [\to 4b + 8c = 8]$ and $9b + 9^{\frac{3}{2}}c = 13.5\ [\to 9b + 27c = 13.5]$; State $b = 3$ and $c = -0.5$. OR $3\times4 + (-0.5)\times4^{\frac{3}{2}} = 8$ AND $3\times9 + (-0.5)\times9^{\frac{3}{2}} = 13.5$ | B1 | Must have 2 correct equations, which do not have to be simplified. Allow to just state the values of $b$ and $c$. Allow substitution of $b=3$ and $c=-0.5$ in both equations to verify. No further calculation required. B0 if any incorrect work seen. |
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## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(a = \dfrac{dv}{dt} =\right) 3 - 0.5\times\dfrac{3}{2}\times t^{\frac{1}{2}}$ OR $b + c\times\dfrac{3}{2}\times t^{\frac{1}{2}}$ | M1 | Attempt to differentiate, decrease power by 1 and a change in coefficient in at least one term (which must be the same term); allow unsimplified; $a = \dfrac{v}{t}$ is M0. |
| acceleration $= 2.25\ \text{ms}^{-2}$ | A1 | OE, e.g. $\dfrac{9}{4}$ or $2\dfrac{1}{4}$. **SC B1** for 2.25 if no differentiation seen. |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $v$ to 0 and attempt to solve for $t$ | M1 | Must get to $t = \ldots$ and must be positive. |
| $t = 36$ ONLY | A1 | WWW. Allow if $t = 0$ seen and not rejected. |
| Attempt to integrate | M1 | Increase power by 1 and a change in coefficient in at least one term (which must be the same term). $s = vt$ is M0. |
| $s = \dfrac{3}{2}t^2 - 0.5\times\dfrac{2}{5}\times t^{\frac{5}{2}}(+D) = \dfrac{3}{2}t^2 - \dfrac{1}{5}t^{\frac{5}{2}}(+D)$ OR $s = \dfrac{b}{2}t^2 + c\times\dfrac{2}{5}\times t^{\frac{5}{2}}(+D)$ | A1 | Allow unsimplified (including indices). |
| Sub $t = 36$ (or use limits 36 and 0) to get distance $= 388.8$ m ONLY | A1 | Allow 389 m. If no integration seen for the last 3 marks, allow **SC B1** for 388.8 m. Max M1A1B1 for 3/5 marks. |
---
## Question 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{3}{2}t^2 - 0.5\times\dfrac{2}{5}\times t^{\frac{5}{2}} = 0$ | M1 | Equate *their* $s$ (that has come from an integration attempt) to 0 and attempt to solve for $t$. Must get to $t = \ldots$ |
| $t = 56.25$ | A1 | WWW. OE, e.g. $\dfrac{225}{4}$. Allow 2sf or better. Allow any correct unsimplified expression equivalent to $56.25$ e.g. $\dfrac{15^2}{2^2}$ or $7.5^2$. |
| Speed $= 42.2\ \text{ms}^{-1}$ ONLY | A1 | AWRT 42.2. Speed $= -42.2$ is A0. Allow A1 if negative sign dropped without justification. |
---6 A particle $P$ starts at rest and moves in a straight line from a point $O$. At time $t$ s after leaving $O$, the velocity of $P , v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by $v = b t + c t ^ { \frac { 3 } { 2 } }$, where $b$ and $c$ are constants. $P$ has velocity $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 4$ and has velocity $13.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 9$.
\begin{enumerate}[label=(\alph*)]
\item Show that $b = 3$ and $c = - 0.5$.
\item Find the acceleration of $P$ when $t = 1$.
\item Find the positive value of $t$ when $P$ is at instantaneous rest and find the distance of $P$ from $O$ at this instant.
\item Find the speed of $P$ at the instant it returns to $O$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q6 [11]}}